Question

Which of the following configurations is associated with the biggest jump between 2^nd and 3^rd I.E.:
\[\begin{align} & \text{(A) 1}{{\text{s}}^{2}}\text{, 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{2}} \\ & \text{(B) 1}{{\text{s}}^{2}}\text{ ,2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{,3}{{\text{s}}^{1}} \\ & \text{(C) 1}{{\text{s}}^{2}}\text{, 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{,3}{{\text{s}}^{2}} \\ & \text{(D) 1}{{\text{s}}^{2}}\text{, 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{1}} \\ & \\ \end{align}\]

Answer 1

Hint: The biggest jump between second and third I.E. is when removal of the second electron leads to noble gas configuration. The difference in I.E. will be when two electrons are easily removed and the third electron is difficult to remove. To remove an electron from noble gas, higher energy is required.
Complete step by step answer:
- I.E., Ionization energy is the minimum amount of energy required to remove an electron from an atom or ion.
- First I.E. is the amount of energy required to remove the first electron.
- Second I.E. is the amount of energy required to remove a second electron or electron from a unipositive ion.
- High amounts of energy is required to remove an electron from noble gas as they are highly stable.
Consider option c
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$
The 3s orbital is completely filled, removal of an electron from 3s will require high energy.
Electronic configuration after removal of first electron
 $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}$
The 3s orbital is half filled and removal of this electron will require less energy as it will lead to noble gas configuration.
Electronic configuration after removal of second electron: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
Third I.E. will be greater than second I.E. as it requires a high amount of energy to remove an electron from noble gas configuration.
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$
configuration is associated with the biggest jump between 2^nd and 3^rd I.E.
So option C is correct.
Note: Three factors affect ionization energy
1) No. of shells-higher the number of shells, less I.E. required to remove an electron.
So, Removal of electrons from 3s will require less energy than 2s.
2) Nuclear charge:
Higher the nuclear charge, higher the amount of energy required to remove an electron.
For example, more amount of energy will be remove an electron

from $M{{g}^{+2}}$ than $N{{a}^{+}}$
3) shielding effect:
As the shielding effect increases, I.E. increases.
$s>p>d>f$